Let $f: G \to H$ be a Group homomorphism, and write $f(G)$ for the set $\{ f(g) : g \in G \}$. Then $f(G)$ is a group under the operation inherited from $H$.

Proof

To prove this, we must verify the group axioms. Let $f: G \to H$ be a group homomorphism, and let $e_G, e_H$ be the identities of $G$ and of $H$ respectively. Write $f(G)$ for the image of $G$.

Then $f(G)$ is closed under the operation of $H$: since $f(g) f(h) = f(gh)$, so the result of $H$-multiplying two elements of $f(G)$ is also in $f(G)$.

$e_H$ is the identity for $f(G)$: it is $f(e_G)$, so it does lie in the image, while it acts as the identity because $f(e_G) f(g) = f(e_G g) = f(g)$, and likewise for multiplication on the right.

Inverses exist, by "the inverse of the image is the image of the inverse".

The operation remains associative: this is inherited from $H$.

Therefore, $f(G)$ is a group, and indeed is a subgroup of $H$.